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- January 13th 2009, 07:17 AM #1

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- January 13th 2009, 07:35 AM #2

- January 13th 2009, 08:30 AM #3

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- January 13th 2009, 08:56 AM #4

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Hi

Let f(x) = tan x - x - x^3/3

Then f'(x) = tan²x - x² = (tan x -x)(tan x + x)

For 0 < x < pi/2 tan x > 0 and x > 0 therefore tan x + x > 0

The sign of f'(x) is the same as g(x) = tan x - x

g'(x) = tan²x > 0

g is increasing on [0,pi/2[

g(0) = 0 so g(x) > 0 on [0,pi/2[

Therefore f'(x) > 0 on [0,pi/2[

f(0) = 0

f(x) > 0 on [0,pi/2[